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At a temperature $T \ll T_F$, the quasiparticles will be excited at an average energy of $\epsilon = k_B T$ above the Fermi surface. Therefore, their lifetime will be proportional to $1 / T^2$, meaning that the conductivity of a Fermi Liquid scales as:  \begin{equation}  \sigma_{\mathrm{FL}} \sim \frac{1}{T^2} \Rightleftarrow \Leftrightarrow  \rho \sim T^2 \end{equation}  Before concluding that conductivity scales as $1 / T^2$, there's one element left to show. A non-infinite conductivity requires both a decay time and an energy loss mechanism. In the current form, the energy the quasiparticle loses goes back to the Fermi sea, effectively accelerating it. What leads to energy loss is that fraction of collisions that bring the crystal momentum of the quasiparticle outside the first Brillouin zone, also known as Umklapp scattering\cite{pal2012}. In essence, whenever the Fermi sea accelerates to a crystal momentum that's outside the first Brillouin zone, this crystal momentum gets translated by a lattice translation vector back into the first Brillouin zone. The energy lost by the Fermi sea is then given to the crystal lattice.\footnote{Though there's something I'm still thinking about. What is the mechanism for energy transfer in Umklapp scattering?}